area of one leaf of a rose

Question

Find the area enclosed by one leaf of the rose r=12cos3θ

Answer 1

One leaf is produced when cos(3θ) starts from the origin then comes back to the origin.

 

cos(3θ) is zero when θ = 30° = π/6 and θ = 90° = π/2.

 

dA = ½r²dθ for the infinitesimal area in polar coordinates.

 

 

A = ∫½r²dθ from π/6 to π/2.

 

A = ½·144 ∫cos²(3θ) dθ = 72 ∫[½ + ½cos(6θ)] dθ

 

A = 72 [θ/2 + (1/12)sin(6θ)] evaluated at π/6 to π/2.

 

A = 72 [ π/4 + (1/12)sin(3π) - π/12 - (1/12)sin(π)]

 

A = 72 [π/6] = 12π